Sketch Level Curves for z=0 and z=1: Tips and Tricks for Accurate Plotting

[ronho1234]

sketch the level curve z=(x^2-2y+6)/(3x^2+y) at heights z=0 and z=1

i have already compute the 2 equations for the 2 z values and drawn it in 2d but when it comes to plotting it with the extra z axis i don't know what to do. please help...

 

[Mentallic]

Well if you have your 3-d axes, then at z=0 you simply draw the equation you found (always stick on the plane z=0 and draw it as you would have drawn the equation in 2-d on the x-y axis) then at z=1 draw the second equation.

 

[ehild]

You do not need to draw the z axis. Plot the level curves in 2D on the x,y plane and mark the curves with z=0 and z=1. Think of a map, the level lines are marked with the height.

ehild

 

[ronho1234]

umm okay so if i draw my two equations on a xy plane i get a concave up with intersect 3 for z=0 and for z=1 i get a concave down intersect 2 and roots -root3 and root3... so are you telling just to leave it at that?...

and for a 3d axis because i have situated my x at the bottom the y then z anticlockwise this will mean that my concave up parabola would be flat on the y-axis and my concave down will be like a saddle-like... i just can't see what the final level curve is suppose to look like when i have two separate parabolas

 

[Mentallic]

Yes sorry, as ehild said, your level curves should be drawn in 2d on the x-y plane with a label of which z value it's following.

If you're struggling to figure out what the shape of curve in 3d space is supposed to look like, try solve the equation for y and see if that sheds any light:

z=\frac{x^2-2y+6}{3x^2+y}

3zx^2+3zy=x^2-2y+6

(3z-2)y=(1-3z)x^2+6

y=\frac{(1-3z)x^2+6}{3z-2}

Now what we can gather from this is that for each level curve of z, we will get some parabola of the form ax^2+b for some constants a,b so the parabola will vary from being concave up and concave down, with differing intersections on the y-axis. This would suggest it has the shape of a saddle, yes?

 

[ronho1234]

yes i kinda get what you're saying its just that because i have two separate parabolas on my xyz axis i don't know where the critical point on the saddle would be and where it would connect and how exactly its suppose to look like...

 

[ronho1234]

since because the question asks only for z=0 and z=1 should i just leave 2 separate parabolas on my xyz axes completely separate of should i have them opposite each other on a xy plane labelled z=0 and z=1?

 

[Mentallic]

yes i kinda get what you're saying its just that because i have two separate parabolas on my xyz axis i don't know where the critical point on the saddle would be and where it would connect and how exactly its suppose to look like...

Well can you figure out what value of z the parabola goes from a positive to a negative concavity? Take a look at the function y=f(x,z).

since because the question asks only for z=0 and z=1 should i just leave 2 separate parabolas on my xyz axes completely separate of should i have them opposite each other on a xy plane labelled z=0 and z=1?

On your xyz axes you should have the parabola drawn at z=0, and the other parabola drawn at z=1. This is to be used as a template to help you draw the curve for all z.

For example, if I wanted to draw z=x^2+y^2 then if I take a few level curves, say for z=0,1,2 then I can use those to get a rough idea of what it should look like in 3-d, and it'll turn out as so:

http://www.google.com.au/search?sourceid=chrome&ie=UTF-8&q=z%3Dx%5E2%2By%5E2

 

[ronho1234]

umm so i did the double derivative of y=f(x,z) and i found z= 1/3 so is it at this point that the concavity changes?
i did try to draw some level curves on the xy plane for z=2 and z=3 i found out that they were both concave down, so i would have z=o level curve concave up and z=1 z=2 z=3 concave curve concave down , so if these values were to be plotted on an xyz graph would this mean that my final level curve by a saddle that is rising/steep around the xz plane and going flatter around z=0 where the y-axis is?

sorry if this is going on for a long while i just can't seem to get it...